反转链表II

难度：中等

题目描述

给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。

示例1

输入： head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例2

输入： head = [5], left = 1, right = 1
输出：[5]

题解

由于没有提及修改原链表节点，那么可以直接修改值，具体方法如下

• 快慢指针，慢指针指向left，快指针指向right，将两个值交换
• 将慢指针右移，快指针移动到right - 1，继续交换

当左右指针会面，即可结束，得到最终答案

想法代码

public class ListNode
{public int val;public ListNode next;public ListNode(int val = 0, ListNode next = null){this.val = val; this.next = next;}
}
class Solution
{public static void Main(String[] args){Solution solution = new Solution();ListNode head = new ListNode(1){next = new ListNode(2){next = new ListNode(3){next = new ListNode(4){next = new ListNode(5) }}}};int left = 2;int right = 4;ListNode ans = solution.ReverseBetween(head, left, right);while (ans != null){Console.Write(ans.val + " ");ans = ans.next;}}public ListNode ReverseBetween(ListNode head, int left, int right){if (left == right){return head;}ListNode ans = head;ListNode slow = head;for (int i = 0; i < left - 1; i++){slow = slow.next;}for (int i = left; i < right; i++){ListNode fast = head;for (int j = 0; j < right - 1; j++){fast = fast.next;}int temp = fast.val;fast.val = slow.val;slow.val = temp;slow = slow.next;right--;}return ans;}
}slow = slow.next;right--;}return ans;}
}